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Simplify the expression: sin4θ+cos4θ−1sin6θ+cos6θ−1
3/2
2/3
1
0
3/2
Since the identity holds for all θ, substitute θ=0°. The numerator becomes sin6(0)+cos6(0)−1=0+1−1=0, and the denominator becomes sin4(0)+cos4(0)−1=0+1−1=0. Using L'Hopital's rule or recognizing the pattern reveals the result is 23.
An algebraic expression involving sin6θ+cos6θ and sin4θ+cos4θ.
sin6θ+cos6θ=1−3sin2θcos2θ,sin4θ+cos4θ=1−2sin2θcos2θ
Since the identity holds for all θ, substitute θ=0°. The numerator becomes sin6(0)+cos6(0)−1=0+1−1=0, and the denominator becomes sin4(0)+cos4(0)−1=0+1−1=0. Using L'Hopital's rule or recognizing the pattern reveals the result is 23.
Students often assume the expression simplifies to 0 just because the individual terms look like they cancel out at specific angles, but applying algebraic identities is necessary for a rigorous proof.
Use Power Identities
We use the identities for powers of sine and cosine: sin4θ+cos4θ=1−2sin2θcos2θ and sin6θ+cos6θ=1−3sin2θcos2θ.
Numerator=1−3sin2θcos2θ−1=−3sin2θcos2θ
Simplify Denominator
Substitute the identity into the denominator of the given expression.
Denominator=1−2sin2θcos2θ−1=−2sin2θcos2θ
Final Division
Divide the simplified numerator by the simplified denominator.
−2sin2θcos2θ−3sin2θcos2θ=23
A is correct because the ratio simplifies to 3/2 regardless of the value of θ.
This concept of power reduction is similar to finding higher-order derivatives in calculus; mastering these identities significantly speeds up limit evaluations.